8道不定积分题,没多少分了,就给15,分不在多,有诚则采!1,∫1/(2x^2-1)dx2,∫1/(4x^2+4x-3)dx上面两题的答案好像是同一类型的,看起来很别扭不知道怎么换算的还是说是带公公式?3,∫(tan^2x + tan^4x)dx4,∫

8道不定积分题,没多少分了,就给15,分不在多,有诚则采!1,∫1/(2x^2-1)dx2,∫1/(4x^2+4x-3)dx上面两题的答案好像是同一类型的,看起来很别扭不知道怎么换算的还是说是带公公式?3,∫(tan^2x + tan^4x)dx4,∫
8道不定积分题,没多少分了,就给15,分不在多,有诚则采!
1,∫1/(2x^2-1)dx
2,∫1/(4x^2+4x-3)dx
上面两题的答案好像是同一类型的,看起来很别扭不知道怎么换算的还是说是带公公式?
3,∫(tan^2x + tan^4x)dx
4,∫cosxcos(x/2)dx
5,∫2x+1/(x^2 - 2x + 2)dx
6,∫x^2/根号(4 - x^2)dx
7,∫1/(x^2 * 根号(x^2 - 1))dx
8,∫根号(x^2 - 1)/x dx

8道不定积分题,没多少分了,就给15,分不在多,有诚则采!1,∫1/(2x^2-1)dx2,∫1/(4x^2+4x-3)dx上面两题的答案好像是同一类型的,看起来很别扭不知道怎么换算的还是说是带公公式?3,∫(tan^2x + tan^4x)dx4,∫
∫ 1/(2x² - 1) dx
= ∫ 1/[(√2x - 1)(√2x + 1)] dx
= (1/2√2)∫ [(√2x + 1) - (√2x - 1)]/[(√2x - 1)(√2x + 1)] d(√2x)
= (1/2√2)∫ [1/(√2x - 1) - 1/(√2x + 1)] d(√2x)
= (1/2√2)(ln|√2x - 1| - ln|√2x + 1|) + C
= ln|(√2x - 1)/(√2x + 1)|/(2√2) + C
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∫ 1/(4x² + 4x - 3) dx
= ∫ 1/[(2x - 1)(2x + 3)] dx
= (1/8)∫ [1/(2x - 1) - 1/(2x + 3)] d(2x)
= (1/8)ln|(2x - 1)/(2x + 3)| + C
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∫ (tan²x + tan⁴x) dx
= ∫ tan²x(1 + tan²x) dx
= ∫ tan²x • sec²x dx
= ∫ tan²x dtanx
= (1/3)tan³x + C
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∫ cosx • cos(x/2) dx
= (1/2)∫ [cos(x + x/2) + cos(x - x/2)] dx
= (1/2)∫ [cos(3x/2) + cos(x/2)] dx
= (1/2)[(2/3)sin(3x/2) + 2sin(x/2)] + C
= (1/3)sin(3x/2) + sin(x/2) + C
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∫ (2x + 1)/(x² - 2x + 2) dx
= ∫ (2x - 2)/(x² - 2x + 2) dx + 3∫ dx/(x² - 2x + 2)
= ∫ d(x² - 2x + 2)/(x² - 2x + 2) + 3∫ dx/[(x - 1)² + 1]
= ln|x² - 2x + 2| + 3arctan(x - 1) + C
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∫ x²/√(4 - x²) dx,x = 2sinθ,dx = 2cosθ dθ
= ∫ 4sin²θ/(2cosθ) • 2cosθ dθ
= 2∫ (1 - cos2θ) dθ
= 2θ - 2 • 1/2sin2θ + C
= 2arcsin(x/2) - 2(x/2)[√(4 - x²)/2] + C
= 2arcsin(x/2) - (x/2)√(4 - x²) + C
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∫ 1/[x²√(x² - 1)] dx,x = secθ,dx = secθtanθ dθ
= ∫ secθtanθ/(sec²θ • tanθ) dθ
= ∫ cosθ dθ
= sinθ + C
= √(x² - 1)/x + C
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∫ √(x² - 1)/x dx,x = secθ,dx = secθtanθ dθ
= ∫ tanθ/secθ • secθtanθ dθ
= ∫ tan²θ dθ
= ∫ (sec²θ - 1) dθ
= tanθ - θ + C
= √(x² - 1) - arcsecx + C
= √(x² - 1) - arccos(1/x) + C